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I face the following problem but disagree with the solution that has been provided to me:

You throw two times a loaded die such that $P(1)=P(3)=P(4)=P(5)=1/8$ and >$P(2)=P(6)=1/4$.

One of the dice gave 6 (but we don't know which one). What is the probability that the sum that you obtain is strictly greater than 10?

The solution that I received is the following:

The results can be written as $(6, x)$ and $(x, 6)$, where $x$ must be equal to 5 or 6. So $P(> 10) = 2×(1/8+1/4)−1/4 = 1/2$.

But I am skeptical about this answer, as the "method" does not work if we try to compute the complement $P(=10$ or lower than $10)$, as the result would give a probability greater than 1...

I however cannot figure out where the problem comes from!

Bardamu
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1 Answers1

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There are $11$ possible results in which one die is a $6$. In three of those cases, the sum is over $10$. We can check the answer by calculating each case individually. We can abbreviate our work somewhat by noting that $P(x,6)=P(6,x)$ for all $x$.

$$\frac{2P(5,6)+P(6,6)}{2P(1,6)+2P(2,6)+2P(3,6)+2P(4,6)+2P(5,6)+P(6,6)}$$

which equals:

$$\frac{\frac{2}{32}+\frac{1}{16}}{\frac{2}{32}+\frac{2}{16}+\frac{2}{32}+\frac{2}{32}+\frac{2}{32}+\frac{1}{16}}=\frac{\frac18}{\frac7{16}}=\frac27$$

The solution you were given is.... a little unclear. It appears someone was calculating the probability that $D1=5$ or $6$ OR $D2=5$ or $6$. It's still not clear where the $-\frac14$ comes from, though. This is a conditional probability question, though, where the given condition is not so simple. I'm trying to find an approach similar to that one, but no luck so far...

G Tony Jacobs
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